9 Calculations from
chemical equations
Answers to end-of-chapter questions
Pages 163–164 Questions
1 molar mass of Fe 2 (SO 4 ) 3 = (2 × 55.8) + 3 × [32.1 + (4 × 16.0)] = 399.9 g mol
molar mass of Fe(OH) 3 = 55.8 + (3 × 17.0) = 106.8 g mol
–1
–1
−1
amount of Fe 2 (SO 4 ) 3 (in moles) = mass/molar mass = 12.7 g/399.9 g mol = 0.03176 mol
moles Fe(OH) 3 = moles Fe 2 (SO 4 ) 3 × number of Fe(OH) 3 in equation/number of Fe 2 (SO 4 ) 3 in
equation
= 0.03176 mol × 2 = 0.06352 mol
–1
mass of Fe(OH) 3 = moles × molar mass = 0.06352 mol × 106.8 g mol = 6.78 g
[e] It is a good idea to work out the molar masses of the two substances involved before starting the
rest of the calculation.
2 molar mass of AgNO 3 = 107.9 + 14.0 + (3 × 16.0) = 169.9 g mol
–1
−1
amount (in moles) of AgNO 3 = mass/molar mass = 12.6 g/169.9 g mol = 0.07416 mol
ratio of Cu:AgNO 3 = 1:2
so, amount of copper = ½ × 0.07416 = 0.03708 mol
mass of copper needed = moles × molar mass
–1
= 0.03708 mol × 63.5 g mol = 2.35 g
[e] The answer 2.35 g was obtained keeping all the numbers on the calculator during the calculation.
If the rounded up value of 0.0371 mol is used, the answer 2.36 g is obtained. Either would score full
marks.
−1
A common error is to calculate the molar mass of AgNO 3 as 2 × 170 g mol because there are
2 moles of it in the equation. This is wrong. The number of moles of a substance depends only on its
mass and not on the reaction. The stoichiometry comes into play when moles of one substance are
converted to moles of another substance. This type of calculation can be done either by mass ratio or
by converting to moles, then using the reaction stoichiometry and finally converting back to mass.
The second method fits all types of calculation, so is the better one to use.
3 molar mass of NaNO 3 = 23.0 + 14.0 + (3 × 16.0) = 85.0 g mol
–1
−1
amount of NaNO 3 = mass/molar mass = 33.3 g/85.0 g mol = 0.392 mol
ratio of O 2 :NaNO 3 = 1:2
so, amount of oxygen = ½ × 0.392 = 0.196 mol
volume of oxygen = moles × molar volume
3
= 0.196 mol × 25.0 dm mol
= 4.90 dm
–1
3
4 ratio of NH 3 :H 2 = 2:3
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9 Calculations from
chemical equations
Answers to end-of-chapter questions
3
so, theoretical amount of ammonia produced = ⅔ × 60 = 40 dm
percentage yield = actual yield of product × 100/theoretical yield of product
percentage yield = 25 × 100/40 = 62.5%
[e] When calculating the percentage yield, the yields can be expressed in moles or in mass units.
Calculating both yields in moles gives:
–1
actual amount (moles) of NH 3 = 2550 g/17.0 g mol = 150 mol
so, percentage yield = 150 mol × 100/1000mol = 15.0%
The percentage yield is not mass of product × 100/mass of reactant.
5 amount (in moles) = concentration × volume
–3
3
= 0.0545 mol dm × 23.4/1000 dm = 0.00128 mol
−3
3
[e] Be careful about units. The concentration is in mol dm , but the volume is in cm . The volume in
3
3
cm must be divided by 1000 to convert it into a volume in dm .
−3
3
3
6 volume = moles/concentration = 0.00164 mol/0.106 mol dm = 0.0155 dm = 15.5 cm
3
3
[e] Volumes of solutions less than 1 dm are usually expressed in cm .
7 amount (moles) of H 2 C 2 O 4 .2H 2 O = concentration × volume
–3
3
= 0.0500 mol dm × 250/1000 dm = 0.0125 mol
molar mass of H 2 C 2 O 4 .2H 2 O = 2.0 + (2 × 12.0) + (4 × 16.0) + (2 × 18.0) = 126.0 g mol
–1
mass of hydrated ethanedioic acid needed = moles × molar mass
–1
= 0.0125 mol × 126.0 g mol = 1.575 g
−1
8 a) amount (moles) of Fe = mass/molar mass = 4.50 g/55.8 g mol = 0.0806 mol
amount of copper(II) sulfate = concentration × volume in dm
–3
3
3
= 2.00 mol dm × 0.0500 dm = 0.100 mol
As 0.100 is greater than 0.0806, iron is the limiting reagent.
[e] As the substances react in a 1:1 ratio, the one with fewer moles is the limiting reagent.
b) ratio of Cu to Fe in the equation = 1:1
so, amount (moles) of copper = amount of iron = 0.0806 mol
–1
mass of copper = moles × molar mass = 0.0806 mol × 63.5 g mol = 5.12 g
c) Copper sulfate is in excess, so the solution will still be blue.
[e] If copper sulfate had been the limiting reagent, the solution would be colourless at the end of the
reaction.
9 a) 2NaOH + H 2 SO 4 → Na 2 SO 4 + 2H 2 O
[e] The stoichiometric ratio is not 1:1, so the amount of sodium sulfate that would be produced from
both sodium hydroxide and sulfuric acid reacting completely has to be calculated. The limiting
reagent is the one that gives the smaller amount of product.
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9 Calculations from
chemical equations
Answers to end-of-chapter questions
b) molar mass of sodium hydroxide = 23.0 + 16.0 + 1.0 = 40.0 g mol
–1
–1
amount (moles) of NaOH = mass/molar mass = 21.5 g/40.0 g mol = 0.5375 mol
amount (moles) of Na 2 SO 4 that would be produced if NaOH were the limiting reagent
= ½ × 0.5375 = 0.269 mol
3
amount (moles) of H 2 SO 4 = concentration × volume in dm
–3
3
= 1.00 mol dm × 0.500 dm = 0.500 mol
amount (moles) of Na 2 SO 4 that would be produced if H 2 SO 4 were the limiting reagent
= 0.500 mol, which is more than 0.269 mol from NaOH
Therefore, sodium hydroxide is the limiting reagent.
c) molar mass of Na 2 SO 4 = (2 × 23.0) + 32.1 + (4 × 16.0) = 142.1 g mol
–1
mass of sodium sulfate produced = moles × molar mass
–1
= 0.269 mol × 142.1 g mol = 38.2 g
d) Sodium hydroxide is the limiting reagent, so there will be an excess of sulfuric acid. Therefore,
red litmus will stay red and blue litmus will turn red.
[e] Even though there are more moles of sodium hydroxide than sulfuric acid, the sodium hydroxide
is the limiting reagent because 2 moles of it are needed for every mole of sulfuric acid.
–6
3
3
3
–1
10 There are 3.13 × 10 dm of CO in 1 dm of air. Assume that the molar volume = 24 dm mol .
–6
–7
amount of CO = 3.13 × 10 /24 = 1.304 × 10 mol
–7
–1
–6
mass of CO = 1.304 × 10 mol × 28 g mol = 3.15 × 10 g (= 3.15 μg)
11 mass of water = 5.46 − 2.98 = 2.48 g
–1
moles of water = 2.48 g/18 g mol = 0.1378 mol
–1
M r of FeSO 4 = 151.9 g mol
moles of FeSO 4 = 2.98/151.9 = 0.01962 mol = moles of hydrated FeSO 4
ratio moles of water:moles of hydrated FeSO 4 = 0.1378:0.01962 = 7.02
number of molecules of water of crystallisation = 7
[e] The answer must be a whole number.
−3
−18
12 a) mass of glucose = 0.054 g dm × 4.0 × 10
−19
moles of glucose = 2.16 × 10
= 2.16 × 10
−19
g
–1
−21
mol
g/180 g mol = 1.2 × 10
−21
b) number of molecules = 1.2 × 10
23
–1
mol × 6.02 × 10 mol = 722
13 a) Error 1: It should be dissolved in a smaller amount of water and the solution and washings
3
3
made up to 250 cm (not dissolved in 250 cm of water).
Error 2: A standard flask should have been rather than a beaker.
3
–1
b) moles of Na 2 CO 3 .H 2 O in 250 cm = 1.55 g/124 g mol = 0.0125 mol
3
moles of Na 2 CO 3 .H 2 O in 25 cm = 0.0125 × 25.0/250 = 0.00125 mol
moles of HCl = 2 × 0.00125 = 0.0025 mol
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9 Calculations from
chemical equations
Answers to end-of-chapter questions
3
−3
[HCl] = 0.0025 mol/0.02670 dm = 0.0936 mol dm
14 ppm is the mass per million g of solvent.
3
−3
−5
moles of HCl = 0.00755 dm × 0.0050 mol dm = 3.775 × 10 mol = mol of KOH
–1
mass of KOH = 3.775 mol × 56.1 g mol = 0.00212 g in 100 g
6
mass of KOH in 10 g of water = 21.2 g = ppm of KOH
Pages 165–166 Exam practice questions
3
1 a) volume of CO 2 = 0.289 dm ()
3
3
−1
moles of CO 2 = 0.289 dm /24.4 dm mol = 0.01184 = moles CaCO 3 ()
mass of CaCO 3 = moles × molar mass = 0.01184 × 100.1 = 1.186 g ()
purity = 100 × 1.186/1.23 = 96.4% ()
b) moles CaCO 3 = moles CO 2 = 0.01184
moles of HCl needed = 2 × 0.0184 = 0.02368 ()
3
3
volume = moles/concentration = 0.02369/2 = 0.01184 dm = 11.8 cm ()
2 a) A ()
b) D ()
[e] The error is ±2 × 0.05 = ±0.1, so % error = 100 × 0.1/11.21 = 0.89%
3
−3
c) moles HCl = 0.02335 dm × 0.110 mol dm = 0.002569 mol ()
3
d) moles NaOH in 25 cm = 0.002569 mol ()
moles in excess = 10 × 0.002569 = 0.02569 mol ()
e) initial moles of NaOH = 0.050 × 1 = 0.050 mol ()
moles reacted with ibuprofen = 0.050 – 0.02569 = 0.02431 mol ()
f) moles of ibuprofen = 0.02431
−1
mass of ibuprofen in 25 tablets = 0.002431 mol × 228 g mol = 5.54 g ()
mass in one tablet = 5.54/25 = 0.222 g = 222 mg ()
3 a) i)
moles = mass/molar mass ()
−1
[e] A good way is to check units: g/g mol = mol
ii)
moles = volume/molar volume ()
iii)
moles = number of particles/Avogadro constant ()
b) D ()
c) D ()
d) A ()
[e] moles Rb = 8.55/85.5 = 0.1; moles O 2 = (11.75 – 8.55)/32 = 0.1; ratio Rb:O 2 = 1:1
e) B ()
© Hodder & Stoughton Limited 2015
9 Calculations from
chemical equations
Answers to end-of-chapter questions
[e] moles CuO = 0.1 = theoretical moles CuSO 4 .5H 2 O
theoretical mass = 0.1 × 249.6 g; % = 100 × 23/24.96 = 92%
4 a) Error 1: not rinsing glass rod () — some sodium carbonate left on rod ()
Error 2: not rinsing beaker () — some sodium carbonate left in beaker ()
3
Error 3: not shaking graduated flask () — so variable concentrations of 25 cm samples ()
b) i)
first titre 25.05, second 25.35, third 25.40 ()
[e] All must be given to 2 decimal places.
ii)
−1
moles Na 2 CO 3 .10H 2 O = 7.03 g/286 g mol = 0.02458 mol ()
3
moles in 25 cm sample = 0.002458 mol ()
moles HCl = 2 × 0.002458 = 0.004916 mol ()
3
−3
[HCl] = 0.004916 mol/0.025375 dm = 0.194 mol dm ()
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